A recent Question asked for all positive integer solutions of a simple quartic in four unknowns:
$$ wxyz = (w+x+y+z)^2 \tag{1}$$
whose satisfaction is necessary for the integer side lengths $a,b,c,d$ of a cyclic quadrilateral having area equal to perimeter, where:
$$ a+b+c+d = 2s = w+x+y+z $$
$$ w = s-a, x = s-b, y = s-c, z = s-d $$
Infinitely many positive integer solutions exist to (1), as Tito Piezas and Will Jagy showed. Will Jagy further showed that by permuting unknowns and Vieta jumping, all such solutions may be reduced to nine Hurwitz fundamental solutions satisfying:
$$ w \ge x \ge y \ge z, \; \; \; xyz \ge 2(w+x+y+z) $$
Show that these nine solutions, up to a permutation of unknowns, are the only ones which satisfy $x,y,z \le w \le x+y+z$.
If $w = x+y+z$, the correspondence $w = s-a$ implies $a=0$ but $b,c,d \gt 0$. Therefore five of the nine fundamental solutions correspond to triangles (degenerate cyclic quadrilaterals). The other four fundamental solutions correspond to (nondegenerate) cyclic quadrilaterals, $a,b,c,d \gt 0$. For more details, see this CW post.
Let $p=xyz,\ t=x+y+z$ so that the Hurwitz condition is $p \ge 2(w+t).$ In the post by Will Jagy, combined with another post by leshik on the same topic, the solutions which satisfy this Hurwitz condition were shown to be precisely the nine solutions you refer to as "these nine solutions".
The equation being solved is, using the $p,t$ definitions, $$wp=(w+t)^2.\tag{1}$$ So as noted in the comment by hardmath above, if one can show that $w \le t$ is equivalent to the Hurwitz condition $p \ge 2(w+t)$ under the condition that $(1)$ holds, then we have what is required.
The Hurwitz condition is equivalent, on multiplication by $w$, to $wp \ge 2w^2+2wt.$ In turn this is equivalent, on using $(1)$, to each of $$(w+t)^2 \ge 2w^2+2wt, \\ w^2+2wt+t^2 \ge 2w^2+2wt, \\ w^2 \le t^2, \\ w \le t $$ All the steps here are reversible, so it seems the Hurwitz condition is equivalent to $x+y+z \ge w$.
This answer seems too simple... if it's way off let me know.